|
| |
|
|
A002162
|
|
Decimal expansion of the natural logarithm of 2.
(Formerly M4074 N1689)
|
|
73
|
|
|
|
6, 9, 3, 1, 4, 7, 1, 8, 0, 5, 5, 9, 9, 4, 5, 3, 0, 9, 4, 1, 7, 2, 3, 2, 1, 2, 1, 4, 5, 8, 1, 7, 6, 5, 6, 8, 0, 7, 5, 5, 0, 0, 1, 3, 4, 3, 6, 0, 2, 5, 5, 2, 5, 4, 1, 2, 0, 6, 8, 0, 0, 0, 9, 4, 9, 3, 3, 9, 3, 6, 2, 1, 9, 6, 9, 6, 9, 4, 7, 1, 5, 6, 0, 5, 8, 6, 3, 3, 2, 6, 9, 9, 6, 4, 1, 8, 6, 8, 7
(list;
constant;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,1
|
|
|
COMMENTS
|
Newton calculated the first 16 terms of this sequence.
|
|
|
REFERENCES
|
S. R. Finch, Mathematical Constants, Cambridge, 2003, Sections 1.3.3 and 6.2.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
|
|
|
LINKS
|
Harry J. Smith, Table of n, a(n) for n = 0..20000
D. H. Bailey and J. M. Borwein, Experimental Mathematics: Examples, Methods and Implications
P. Bala, New series for old functions
Paul Cooijmans, Odds.
X. Gourdon and P. Sebah, The logarithm constant:log(2)
I. Newton, The method of fluxions and infinite series; with its application to the geometry of curve-lines, 1736; see p. 96.
Simon Plouffe, log(2), natural logarithm of 2 to 2000 places
S. Ramanujan, Question 260, J. Ind. Math. Soc.
D. W. Sweeney, On the computation of Euler's constant, Math. Comp., 17 (1963), 170-178.
Horace S. Uhler, Recalculation and extension of the modulus and of the logarithms of 2, 3, 5, 7 and 17, Proc. Nat. Acad. Sci. U. S. A. 26, (1940). 205-212.
Eric Weisstein's World of Mathematics, Natural Logarithm of 2, Masser-Gramain Constant, Logarithmic Integral
Wikipedia, Natural logarithm of 2.
|
|
|
FORMULA
|
log(2) = Sum_{k>=1} 1/(k*2^k) = Sum_{j>=1} (-1)^(j+1)/j.
log(2) = Integral_{t=0..1} dt/(1+t).
log(2) = 2/3 * (1 + sum(k>=1, 2/[(4k)^3-4k] )) (Ramanujan).
log(2) = 4*sum_{k>=0} [3-2*sqrt(2)]^(2k+1)/(2k+1) (Y. Luke). - R. J. Mathar, Jul 13 2006
log(2) = 1-(1/2)*sum_{k>=1} 1/(k*(2k+1)). - Jaume Oliver Lafont, Jan 06 2009, Jan 08 2009
log(2) = 4*sum_{k>=0} 1/((4k+1)(4k+2)(4k+3)). - Jaume Oliver Lafont, Jan 08 2009
log(2) = 7/12+24*sum_{k>=1} 1/(A052787(k+4)*A000079(k)). - R. J. Mathar, Jan 23 2009
From Alexander R. Povolotsky, Jul 04 2009: (Start)
log(2) = 1/4*(3 - sum(n>=1, 1/(n*(n+1)*(2*n+1)) )).
log(2) = (230166911/9240-sum(k>=1, (1/2)^k* (11/k+10/(k+1)+9/(k+2)+8/(k+3)+7/(k+4)+6/(k+5)-6/(k+7)-7/(k+8)-8/(k+9) -9/(k+10)-10/(k+11)) ))/35917. (End)
log(2) = A052882/A000670. - Mats Granvik, Aug 10 2009
From log(1-x-x^2) at x=1/2, log(2)=(1/2)*Sum_{k>=1}L(k)/(k*2^k), where L(n) is the n-th Lucas number (A000032). - Jaume Oliver Lafont, Oct 24 2009
log(2) = Sum_{k>=1} 1/(cos(k*Pi/3)*k*2^k) (Cf. A176900). - Jaume Oliver Lafont, Apr 29 2010
log(2) = (sum(n>=1, 1/(n^2*(n+1)^2*(2*n+1)) ) +11)/16. - Alexander R. Povolotsky, Jan 13 2011
log(2) = sum(n>=1, (2*n+1)/(sum(k=1..n, k^2 )^2)) +396)/576. - Alexander R. Povolotsky, Jan 14 2011
log(2) = 105*(sum(n>=1, 1/(2*n*(2*n+1)*(2*n+3)*(2*n+5)*(2*n+7)) ) - 319/44100) log(2) = (319/420 - 3/2*sum(n>=1, 1/(6*n^2+39*n+63) )). - Alexander R. Povolotsky, Dec 16 2008
log(2) = sum(k>=1, A191907(2,k)/k ). - Mats Granvik, Jun 19 2011
log(2) = Integral_{x=0..oo} 1/(1 + e^x) dx. - Jean-François Alcover, Mar 21 2013
log(2) = limit of zeta(s)*(1-1/2^(s-1)) as s -> 1. - Mats Granvik, Jun 18 2013
From Peter Bala, Dec 10 2013: (Start)
log(2) = 2*sum {n = 1..inf} 1/( n*A008288(n-1,n-1)*A008288(n,n) ), a result due to Burnside.
log(2) = 1/3*sum {n >= 0} (5*n+4)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(1/2)^n = 1/12*sum {n >= 0} (28*n+17)/( (3*n+1)*(3*n+2)*C(3*n,n) )*(-1/4)^n.
log(2) = 3/16*sum {n >= 0} (14*n+11)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(1/4)^n = 1/12*sum {n >= 0} (34*n+25)/( (4*n+1)*(4*n+3)*C(4*n,2*n) )*(-1/18)^n. For more series of this type see the Bala link.
See A142979 for series acceleration formulas for log(2) obtained from the Mercator series log(2) = sum {n >= 1} (-1)^(n+1)/n. See A142992 for series for log(2) related to the root lattice C_n. (End)
log(2) = sum(k=2^n..2^(n+1)-1, 1/k) as n -> Infinity. - Richard R. Forberg, Aug 16 2014
From Peter Bala, Feb 03: (Start)
log(2) = 2/3*Sum {k >= 0} 1/((2*k + 1)*9^k).
Define a pair of integer sequences A(n) = 9^n*(2*n + 1)!/n! and B(n) = A(n)*sum {k = 0..n} 1/((2*k + 1)*9^k). Both satisfy the same second order recurrence equation u(n) = (40*n + 16)*u(n-1) - 36*(2*n - 1)^2*u(n-2). From this observation we obtain the continued fraction expansion log(2) = 2/3*(1 + 2/(54 - 36*3^2/(96 - 36*5^2/(136 - ... - 36*(2*n - 1)^2/((40*n + 16) - ... ))))). Cf. A002391, A073000 and A105531 for similar expansions. (End)
|
|
|
EXAMPLE
|
0.693147180559945309417232121458176568075500134360255254120680009493393...
|
|
|
MATHEMATICA
|
RealDigits[N[Log[2], 200]][[1]] (* Vladimir Joseph Stephan Orlovsky, Feb 21 2011 *)
|
|
|
PROG
|
(PARI) { default(realprecision, 20080); x=10*log(2); for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b002162.txt", n, " ", d)); } /* Harry J. Smith, Apr 21 2009 */
|
|
|
CROSSREFS
|
Cf. A016730 Continued fraction. A008288, A142979, A142992.
Sequence in context: A129938 A022698 A013707 * A072365 A239068 A085138
Adjacent sequences: A002159 A002160 A002161 * A002163 A002164 A002165
|
|
|
KEYWORD
|
cons,nonn
|
|
|
AUTHOR
|
N. J. A. Sloane, Apr 30 1991
|
|
|
STATUS
|
approved
|
| |
|
|
